Solve Systems of Equations by Elimination: example 5.30

Solve the system by elimination. ${\begin{matrix} 3 x + 4 y = 12 \\ y = 3 - \frac{3}{4} x \end{matrix}$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solve the system by elimination. ${\begin{matrix} 5 x - 3 y = 15 \\ y = −5 + \frac{5}{3} x \end{matrix}$

How did it go?

	${\begin{matrix} 3 x + 4 y = 12 \\ y = 3 - \frac{3}{4} x \end{matrix}$
Write the second equation in standard form.	${\begin{matrix} 3 x + 4 y & = & 12 \\ \frac{3}{4} x + y & = & 3 \end{matrix}$
Clear the fractions by multiplying the second equation by 4.	${\begin{matrix} 3 x + 4 y & = & 12 \\ 4 (\frac{3}{4} x + y) & = & 4 (3) \end{matrix}$
Simplify.	${\begin{matrix} 3 x + 4 y & = & 12 \\ 3 x + 4 y & = & 12 \end{matrix}$
To eliminate a variable, we multiply the second equation by $−1$ . Simplify and add.	$\begin{matrix} \underset{________________}{{\begin{matrix} 3 x + 4 y & = & 12 \\ - 3 x - 4 y & = & −12 \end{matrix}} \\ 0 = 0 \end{matrix}$

	${\begin{matrix} 3 x + 4 y = 12 \\ y = 3 - \frac{3}{4} x \end{matrix}$
Write the second equation in standard form.	${\begin{matrix} 3 x + 4 y & = & 12 \\ \frac{3}{4} x + y & = & 3 \end{matrix}$
Clear the fractions by multiplying the second equation by 4.	${\begin{matrix} 3 x + 4 y & = & 12 \\ 4 (\frac{3}{4} x + y) & = & 4 (3) \end{matrix}$
Simplify.	${\begin{matrix} 3 x + 4 y & = & 12 \\ 3 x + 4 y & = & 12 \end{matrix}$
To eliminate a variable, we multiply the second equation by $−1$ . Simplify and add.	$\begin{matrix} \underset{________________}{{\begin{matrix} 3 x + 4 y & = & 12 \\ - 3 x - 4 y & = & −12 \end{matrix}} \\ 0 = 0 \end{matrix}$